BIS 102-Theg-MII

BIS 102-Theg-MII - I... ,4 Your name Page 1 of 9 Name...

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Unformatted text preview: I... ,4 Your name Page 1 of 9 Name Bis102, Fall 2001 Last, First Steven Theg Your Social Security Number: Biological Sciences 102 Second Midterm, November 20, 2001 Instructions: 1. There are eight pages in this exam (including the graph paper). Please count them now to be sure you have them all. 2. Write your name and social security number on the first page, and your name on each subsequent page. 3. Write your answers in the space provided below each question. If you need more space, use the back of the page. For calculation problems, a clear line of calculations must be shown to obtain full credit. Student Authorization I authorize the University to distribute publicly this graded exam (eg, handed out in class or left in a bin for me to pick up). Signature , Date Your name Page 2 of 9 1. (4 pts.) \ \ Here are the equations describing the different inhibitor kinetics about which we talked in class during this last examination period. Please match the equation with the statements by placing a letter on the line corresponding to the appropriate equation. A Z A + Kd A. This equation describes the accumulation of a product as a function of time in a first order reaction. B B. P(t) = 50(1—e‘k') — —~ B. This equation describes the A. Y _ Vmax [5] relationship between a first order rate C' V” _ . K constant and ener of acti ation v . [mg] [51+ M gy I F K’ C. This equation describes competitive inhibition. E D Y — A" _ A" + Kd D. This is the Michaelis-Menten equation. H V - . . . " E. 120 2 “ML E. This equation describes [S] + K [I + H1) uncompetitive inhibition. “C M K] F. This equation describes the binding of a single ligand to a protein. F. k = k—Texp[ E”) 'G. This equation describes noncompetitive inhibition. _G_ Vmax [S] G v0 = —— o —— . . . . [I] ([S] + K M ) H. This equation describes cooperative 1+ F 1 binding of ligands to a protein. 1 —D— H V0 _ V"... '[S] Your name Page 3 0f9 2. (6 pts.) Draw the chemical reaction scheme involving enzyme (E), substrate (S), inhibitor (1) and product (P) that forms the starting point for the derivation of the relationship between v0 and [S] in the presence of a non—competitive inhibitor. Start with the scheme below, add the inhibitor and complete it. E + S ES E + P k4 +1 + I H H ———) EI EIS (—— + S 3. (4 pts.) Answer the following questions about the mechanism of action of acetylcholine esterase. What three amino acids comprise the triad? serine, histidine, aspartate The two products of the breakdown of acetylcholine are choline and acetic acid . Of these, which comes off first? Choline Why is it said that acetylcholine esterase operates by a charge relay mechanism? That is, what charge is relayed and where? A proton is relayed from S to H and toward D. This enzyme is inhibited under mildly acidic conditions. Why? The histidine must be deprotonated to withdraw the proton from serine, and this happens above pH 6.0 Your name Page 4 of 9 4. (20 pts.) The table below shows data collected for an enzyme reaction that was run with and without an inhibitor, 1, at 14 mM. Using the empty spaces in the table and the graph paper on the next page, determine Vmax, KM, Vmampp, and KMqapp. What kind of inhibitor is 1? What is the value of K1? 0 You must report the units on these parameters for full credit. 0 You must show your work to receive full credit. Vmax = 150 umolls Vmax,app = 150 umol/s The type of inhibition observed : competitive KI=10mM - Here's a tip: Place the zero point on the abscissa (X-axis) at the middle of the page. 0 Here's another: Do you need to plot all these points? (( Your name Page 6 of 9 5 (15 pts.) You are running a research project looking at three different enzymes. You have given one enzyme to each of three graduate students and asked them to determine the respective kcat's and KM's for the favored substrates. At the week's end you have a lab meeting and are given the following results: Alice's enzyme: kw = 2,000 s“; KM 2 4 x 10‘5 M Frank's enzyme: km = 600 s“; KM = 3 x 10'6 M John's enzyme: kw : 40,000 s"; KM = 8 x 10'7 M Who's enzyme has the highest turnover number? John’s (or since John made the mistake, Alice’s) Who's enzyme has the highest affinity for its substrate? John’s (or since John made the mistake, Frank’s) Which one of these students has made a mistake somewhere? John How do you know? The kcat/KM value for John’s enzyme is impossibly high, beyond the diffusion controlled limit. Of the two remaining, who’s enzyme would you consider the most efficient? That is, who's enzyme is the least likely to show improvement by any possible changes in structure? Frank’s Why? His enzyme is already operating at the diffusion controlled limit 6. (7 pts.) Derive an expression for v0 in which kcat/KM is a second order rate constant. HINT: You might start by thinking about the idea that KM reports the affinity of the enzyme for the substrate, and how that came up. Given the steady-state assumption, v0 = kmx [ES]. Since KM reports the affinity of E for S, it can be thought of as the dissociation constant of the ES complex, .i.e., A, [E][S] [S][E] : [SHE] ES E+S,and KM=Kd= ,so[ES]: ‘—kl [ES] Kcl KM kCfli then v0 = K [E][S] M Your name Page ' ———_—__ 7. (6 pts.) For the following enzyme catalyzed reaction: unaltered substrate product released which one of the following is the most likely shape for the enzyme’s active site (circle the letter)? Justify your answer (briefly, one sentence). a b c arr—a The Bart Simpson enzyme The enzyme’s active site should reflect the shape of the transition complex, which is d. S. (10 pts.) In your analysis on a new protein you find: a gel filtration chromatography results in a peak eluting at 300 kDa. o Non-reducing PAGE with SDS gives two bands at 90 and 60 kDa, respectively. - Reducing PAGE with 303 gives bands at 50, 4S and ID kDa, respectively. What can you say about the structure ofthis protein? Why do you say that? The protein consists of three subunits — I’ll call them or (50 kDa)1 B (45 kDa) and 'y (10 kDa). Two [3 subunits are held together by one or more disulfides to form a larger subunit, call it A, at 90 kDa. Another larger subunit at 60 kDa, call it B, is formed by an or-y dimer held together by one or more disulfides. The structure if the final 300 kDa protein is A233, or Cl: B472- Your name Page 8 of 9 9. (14 pts.) A. Cooperative binding of a ligand to a protein like hemoglobin is characterized by a unique shape of the binding curve (plotting fraction of sites occupied against the concentration of the ligand). On the graph below, sketch the binding curve for a protein with four binding sites when: a. The binding is totally non—cooperative. b. The binding is completely cooperative. c. The binding has some, but not complete, cooperative nature When sketching the curves, their relative shapes are important, not the absolute numbers they represent. Be sure to label the individual curves clearly. {Si B. Write the general form of the equation that describes cooperative binding. What have you put into this equation that accounts for the cooperativity? (TWO ANSWERS HERE) An Y : —— A“ + K d The parameter n confers cooperativity. 10. (14) , When an enzyme with a KM = 0.5 mM is assayed at [S] = 75 mM, an initial rate of 2.5 uM/min is recorded. If you were to assay this same enzyme at [S] = 2 uM, how much product would be formed after 3 5? Because 75 mM is so high compared to KM, the condition [S] >> KM applies and v0 = Vmax. So, 2.5 uM/min is Vmax. When assayed at [S] = 2 uM, [S] << KM, and Vmax / KM = the first order rate constant. So k = 2.5 uM/min / 500 uM = 0.005 min'l. —0.005><i P = sou-er“ ) = 2(1-e 6° ) = 0.0005 uM = 0.5 nM ...
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This note was uploaded on 01/31/2012 for the course BIS 102 taught by Professor Hilt during the Winter '08 term at UC Davis.

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BIS 102-Theg-MII - I... ,4 Your name Page 1 of 9 Name...

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